A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
FB = qv  B =  -e  viˆ  Bjˆ = -evBkˆ
 FB
FE
E
v
x
FB = -e vB = evB
FB =FE
 evB = eE
FE must be in the +z direction,
so E must be in the –z
direction (because the charge
on the electron is negative.
FE = qE = -e E = eE
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
evB =eE
B
y
z
x
 FB
FE
E
v
E = vB
1
2K
2
me v =K  v =
2
me
2K
E =B
me
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
 FB
FE
E
x
electron mass is
on OSE sheet
v
2K
E =B
me
E =  0.015 
don’t forget to
convert eV to J
2  750 1.6 10-19 
9.11 10-31
E =  0.015  1.623 107 
5 N
E = 2.43 10
C
this is speed in
m/s; getting close
to relativistic
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
 FB
FE
E
Problem asks for magnitude
and direction, so
v
N
, in the - z direction
C
Also “legal:”
x
E = 2.43 105
E = 2.43 105

5 N ˆ
E = -  2.43 10  k
C

N
, into page
C
E = 2.43 105
N
,
C